1. CPSC 668Set 5: Synchronous LE in Rings1 CPSC 668 Distributed Algorithms and Systems Spring 2008 Prof. Jennifer Welch

2. CPSC 668Set 5: Synchronous LE in Rings2 Leader Election in Synchronous Rings Here is a simple algorithm. Group rounds into phases, each phase containing n rounds In phase i, the processor with id i, if there is one, sends a message around the ring and is elected.

3. CPSC 668Set 5: Synchronous LE in Rings3 Example of Simple Synchronous Algorithm n = 4, the smallest id is 7. In phases 0 through 6 (corresponding to rounds 1 through 28), no message is ever sent. At beginning of phase 7 (round 29), processor with id 7 sends message which is forwarded around the ring.

4. CPSC 668Set 5: Synchronous LE in Rings4 Analysis of Simple Algorithm Correctness: Easy to see. Message complexity: O(n), which is optimal Time complexity: O(n*m), where m is the smallest id in the ring. –not bounded by any function of n!

5. CPSC 668Set 5: Synchronous LE in Rings5 Another Synchronous Algorithm Works in a slightly weaker model than the previous synchronous algorithm: –processors might not all start at same round; a processor either wakes up spontaneously or when it first gets a message –uniform (does not rely on knowing n)

6. CPSC 668Set 5: Synchronous LE in Rings6 Another Synchronous Algorithm A processor that wakes up spontaneously is active; sends its id in a fast message (one edge per round) A processor that wakes up when receiving a msg is relay; never in the competition A fast message carrying id m becomes slow if it reaches an active processor; starts traveling at one edge per 2 m rounds A processor only forwards a msg whose id is smaller than any id is has previously sent If a proc. gets its own id back, elects self

7. CPSC 668Set 5: Synchronous LE in Rings7 Analysis of Synchronous Algorithm Correctness: convince yourself that the active processor with smallest id is elected. Message complexity: Winner's msg is the fastest. While it traverses the ring, other msgs are slower, so they are overtaken and stopped before too many messages are sent.

8. CPSC 668Set 5: Synchronous LE in Rings8 Message Complexity Divide msgs into three kinds: 1.fast msgs 2.slow msgs sent while the leader's msg is fast 3.slow msgs sent while the leader's msg is slow Next, count the number of each type of msg.

9. CPSC 668Set 5: Synchronous LE in Rings9 Show that no processor forwards more than one fast msg: Suppose p i forwards p j 's fast msg and p k 's fast msg. When p k 's fast msg arrives at p j : –either p j has already sent its fast msg, so p k 's msg becomes slow, or –p j has not already sent its fast msg, so it never will Number of type 1 msgs is n. Number of Type 1 Messages pkpk pjpj pipi

10. CPSC 668Set 5: Synchronous LE in Rings10 Number of Type 2 Messages (slow sent while leader's msg is fast) Leader's msg is fast for at most n rounds –by then it would have returned to leader Slow msg i is forwarded n/2 i times in n rounds Max. number of msgs is when ids are small as possible (0 to n-1 and leader is 0) Number of type 2 msgs is at most ∑n/2 i ≤ n i=1 n-1

11. CPSC 668Set 5: Synchronous LE in Rings11 Number of Type 3 Messages (slow msgs sent while leader's is slow) Maximum number of rounds during which leader's msg is slow is n*2 L (L is leader's id). No msgs are sent once leader's msg has returned to leader Slow msg i is forwarded n*2 L /2 i times during n*2 L rounds. Worst case is when ids are L to L + n-1 Number of type 3 msgs is at most ∑n*2 L /2 i ≤ 2n i=L L+n-1

12. CPSC 668Set 5: Synchronous LE in Rings12 Total Number of Messages We showed –number of type 1 msgs is at most n –number of type 2 msgs is at most n –number of type 3 msgs is at most 2n Thus total number of msgs is at most 4n = O(n).

13. CPSC 668Set 5: Synchronous LE in Rings13 Time Complexity of Synchronous Algorithm Running time is O(n 2 x ), where x is smallest id. Even worse than previous algorithm, which was O(n x) Both algorithms have two potentially undesirable properties: –rely on numeric values of ids to count –number of rounds bears no relationship to n, but depends on minimum id Next result shows that to obtain linear msg complexity, an algorithm must rely on numeric values of the ids.

14. CPSC 668Set 5: Synchronous LE in Rings14 Comparison-Based LE Algorithms We will show that if nodes cannot rely on the numeric values of the ids, then any synchronous LE algorithm has message complexity  (n log n). How do we formalize "not relying on numeric values of ids"? Need a definition of "comparison-based", similar to that for sorting algs.

15. CPSC 668Set 5: Synchronous LE in Rings15 Definition of Comparison-Based First note that in the synchronous model, the behavior of an LE algorithm is totally determined by the distribution of the ids. Denote execution on ring R by exec(R). An LE algorithm is comparison-based if, in any two "order-equivalent" rings R 1 and R 2, "matching" processors p i in R 1 and p j in R 2 have "similar" behaviors in exec(R 1 ) and exec(R 2 ).

16. CPSC 668Set 5: Synchronous LE in Rings16 Definition of Order-Equivalent Rings R 1 = (x 1,x 2,…,x n ) and R 2 = (y 1,y 2,…,y n ) are order-equivalent if x i < x j if and only if y i < y j Example: x1 x1x1 x5x5 x4x4 x3x3 x2x2 y1y1 y2y2 y3y3 y4y4 y5y5

17. CPSC 668Set 5: Synchronous LE in Rings17 Definition of Matching Processors Processors p i in R 1 and p j in R 2 are matching if they are the same distance from the processor with minimum id. Example: p 0 (with id 18) in R 1 and p 1 (with id 5) in R 2, which are 2 hops from p 3 (resp., p 4 )

18. CPSC 668Set 5: Synchronous LE in Rings18 Definition of Similar Behaviors Two processors have similar behaviors in two executions if, in every round –one processor sends a message to the left (right) if and only if the other one does –one processor is elected if and only if the other one is

19. CPSC 668Set 5: Synchronous LE in Rings19 Synchronous Lower Bound Theorem (3.18): For every n ≥ 8 that is a power of 2, there is a ring of size n on which any synchronous comparison-based algorithm sends  (n log n) msgs. Proof: For each n, construct a highly symmetric ring S n on which every comparison-based algorithm takes lots of messages. Symmetry means many processors have order-equivalent neighborhoods and thus behave similarly (including sending msgs)

20. CPSC 668Set 5: Synchronous LE in Rings20 Proof Strategy Define highly symmetric ring S n Show that the number of "active" rounds is at least n/8 Show that in the k-th active round, at least n/(2(2k+1)) msgs are sent. Do some arithmetic to show that ∑ n/(2(2k+1)) =  (n log n) k=1 n/8

21. CPSC 668Set 5: Synchronous LE in Rings21 Active Rounds A round is active if at least one processor sends a msg A proc. in a synchronous algorithm can potentially learn something even in an inactive round –cf. the simple synchronous alg But in a comparison-based alg., a proc. can't learn about the order pattern of its ring in an inactive round Note that the k-th active round might be much later than the k-th round

22. CPSC 668Set 5: Synchronous LE in Rings22 A Highly Symmetric Ring S n Let p i 's id be rev(i), the integer whose binary representation using log n bits is the reverse of i 's binary rep. Example: For technical reasons, then multiply id by n+1 p3p3 p2p2 p1p1 p0p0 2 3 0 1 0 155 10

23. CPSC 668Set 5: Synchronous LE in Rings23 Why Lots of Active Rounds Show number of active rounds, T, in exec(S n ) is at least n/8. Proof: Suppose T < n/8. Let p i be the elected leader. Number of T-neighborhoods order-equivalent to p i 's is at least n/(2(2T+1)) (L. 3.19, to be shown) Since n ≥ 8 and T 1. So there is p j  p i whose T-neighborhood is order- equivalent to p i 's. Then p j is also elected (L. 3.17, to be shown), contradiction.

24. CPSC 668Set 5: Synchronous LE in Rings24 Why Enough Msgs Sent in Each Active Round Show at least n/(2(2k + 1)) msgs are sent in kth active round (L. 3.21) Proof: Since the round is active, at least one proc., say p i, sends a msg. There are at least n/(2(2k+1)) procs. whose k- neighborhood is order-equivalent to p i 's (L. 3.19, to be shown) Each of those procs. sends a msg in kth active round, since p i does (L. 3.17, to be shown).

25. CPSC 668Set 5: Synchronous LE in Rings25 Why So Many O-E Neighborhoods L. 3.19: Show every k-nbrhood of S n has at least n/(2(2k+1)) order-equivalent k-nbrhoods. Proof Sketch: Let N be a k-nbrhood (sequence of 2k+1 ids in the ring). Let j be smallest power of 2 larger than 2k+1. Break S n into n/j segments of length j, with one segment encompassing N. Claim: Each segment is order-equivalent to N.

26. CPSC 668Set 5: Synchronous LE in Rings26 Why Similar Behavior in Order- Equivalent Neighborhoods Intuition is that two nodes in order- equivalent neighborhoods in the same ring will behave similarly, at least until differentiating information from beyond those neighborhoods has reached them. Yet the definition of comparison-based algorithm only requires similar behavior for matching processors in different (albeit order-equivalent) rings.

27. CPSC 668Set 5: Synchronous LE in Rings27 Order-Equivalent Means Similar L. 3.17: Show that if p i and p j have order- equivalent k-nbrhoods, then p i and p j have similar behaviors through k-th active round. Proof Sketch: Construct another ring S n ' such that –p i in S n behaves same as p j in S n ' –p j in S n ' behaves similarly to p j in S n

28. CPSC 668Set 5: Synchronous LE in Rings28 Order-Equivalent Means Similar In more detail, construct S n ' such that –p i 's k-nbrhood in S n equals p i 's in S n ' so behavior is same through k-th active round –S n ' and S n are order-equivalent –p j in S n ' is matching to p j in S n so behavior is similar through k-th active round Can be done since ids in S n are spaced (multiplied by n+1) So p i and p j have similar behavior in S n through k-th active round

29. CPSC 668Set 5: Synchronous LE in Rings29 Anonymous Rings Revisited Leader election is impossible in anonymous rings No way to break symmetry No deterministic algorithm which works in every execution Another way to break symmetry, which works in some (but not all) executions is to use randomization.

30. CPSC 668Set 5: Synchronous LE in Rings30 Randomized Algorithm In each computation step, the processor receives an additional input to its state transition function, a random number.

31. CPSC 668Set 5: Synchronous LE in Rings31 Revised LE Problem Definition Weakened problem definition compared to original: At most one leader is elected in every state of every admissible execution –same as previous definition At least one leader is elected "with high probability". –weaker than previous definition But what does "with high probability" mean?

32. CPSC 668Set 5: Synchronous LE in Rings32 Randomized LE Algorithm Assume synchronous model Initially: –set id to 1 with probability 1 - 1/n and to 2 with probability 1/n –send id to left When msg M is received: –if M contains n ids then if id is unique maximum in M then elect self else not elected –else append id to M and send to left

33. CPSC 668Set 5: Synchronous LE in Rings33 Analysis of Randomized LE Alg. Uses O(n 2 ) msgs There is never more than one leader Sometimes there is no leader –leader is only elected if there is exactly one processor that sets its id to 2 How often is there no leader, i.e., what is the probability? Need some more definitions…

34. CPSC 668Set 5: Synchronous LE in Rings34 Random Choices and Probabilities Since system is synchronous, an admissible execution of the algorithm is determined solely by the initial random choices. Call this collection random choices RC = where each r i is either 1 or 2. Let exec(RC) be the resulting execution. Definition: For any predicate P on executions, Pr[P] is the probability of {RC : exec(RC) satisfies P} I.e., the proportion of random choices resulting in an execution that satisfies P.

35. CPSC 668Set 5: Synchronous LE in Rings35 Probability of Electing a Leader Let P be the predicate "there is a leader". Pr[P] = probability RC contains exactly one 2

36. CPSC 668Set 5: Synchronous LE in Rings36 Improving the Probability of Electing a Leader If procs. notice there is no leader, they can try again. Each iteration of the basic algorithm is a phase. Keep trying until success. Random choices that define an execution consist of an infinite sequence of 1's and 2's, one for each proc. It is possible that algorithm doesn't terminate.

37. CPSC 668Set 5: Synchronous LE in Rings37 Probability of Not Terminating Probability of terminating in a particular phase is (1 - 1/n) n-1 Probability of not terminating in a particular phase is 1 - (1 - 1/n) n-1 Probability of not terminating in k phases is (1 - (1 - 1/n) n-1 ) k since phases are independent. Last expression goes to 0 as k increases.

38. CPSC 668Set 5: Synchronous LE in Rings38 Expected Number of Phases Definition: The expected value of a random variable T is E[T] =  k Pr[T = k] Let T be number of phases until termination. Pr[T = k] = Pr[first k-1 phases fail & kth succeeds] = (1 - (1 - 1/n) n-1 ) k-1 (1 - 1/n) n-1 = (1 - p) k-1 p, where p = (1 - 1/n) n-1 This is a geometric random variable with expected value p -1 < e. So expected number of phases is < 3. k